We'll now take the previous example and change switch 2 (SW2) to a normally closed symbol (load bar instruction). SW1 will be physically OFF and SW2 will be physically ON initially. The ladder diagram now looks like this:
REGISTER 00 | |||||||||||||||
15 | 14 | 13 | 12 | 11 | 10 | 09 | 08 | 07 | 06 | 05 | 04 | 03 | 02 | 01 | 00 |
1 | 0 |
REGISTER 05 | |||||||||||||||
15 | 14 | 13 | 12 | 11 | 10 | 09 | 08 | 07 | 06 | 05 | 04 | 03 | 02 | 01 | 00 |
0 |
*Although most of the items in the register tables above are empty, they should each contain a 0. They were left blank to emphasize the locations we were concerned with.
LOGICAL CONDITION OF SYMBOL | |||
LOGIC BITS | LD | LDB | OUT |
Logic 0 | False | True | False |
Logic 1 | True | False | True |
The plc will only energize an output when all conditions on the rung are TRUE. So, looking at the table above, we see that in the previous example SW1 has to be logic 1 and SW2 must be logic 0. Then and ONLY then will the coil be true (i.e. energized). If any of the instructions on the rung before the output (coil) are false then the output (coil) will be false (not energized).
Let's now look at a truth table of our previous program to further illustrate this important point. Our truth table will show ALL possible combinations of the status of the two inputs.
Inputs | Outputs | Register Logic Bits | |||
SW1(LD) | SW2(LDB) | COIL(OUT) | SW1(LD) | SW2(LDB) | COIL(OUT) |
False | True | False | 0 | 0 | 0 |
False | False | False | 0 | 1 | 0 |
True | True | True | 1 | 0 | 1 |
True | False | False | 1 | 1 | 0 |
CLICK ON PICTURE FOR REGISTER APPLICATION
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